Question: Factor completely. $81x^2+180x+100=$
Both $81x^2$ and $100$ are perfect squares, since $81x^2=({9x})^2$ and $100=({10})^2$. Additionally, $180x$ is twice the product of the roots of $81x^2$ and $100$, since $180x=2({9x})({10})$. $81x^2+180x+100 = ({9x})^2+2({9x})({10})+({10})^2$ So we can use the square of a sum pattern to factor: ${a}^2 +2( a)( b)+ {b}^2 =({a}+{b})^2$ In this case, ${a}={9x}$ and ${b}={10}$ : $ ({9x})^2+2({9x})({10})+({10})^2 =({9x}+{10})^2$ In conclusion, $81x^2+180x+100=(9x +10)^2$ Remember that you can always check your factorization by expanding it.